Question

In Fig. 9, is shown a sector OAP of a circle with centre O, containing $\angle \theta$. AB is perpendicular to the radius OA and meets OP produced at B. Prove that the perimeter of shaded region is

$r[tan\theta +sec\theta +\frac{\pi \theta }{180}-1]$

Answer 1

Given, the radius of circle with centre O is r.

$\angle POA=\theta$.

then, length of the arc $\widehat{PA}=\frac{2\pi r\theta }{{360}^{\circ }}=\frac{\pi r\theta }{{180}^{\circ }}$

And $tan\theta =\frac{AB}{r}$

AB = $rtan\theta$

And $sec\theta =\frac{OB}{r}$

$OB=rsec\theta$

Now, $PB=OB-OP$

= $rsec\theta -r$

$\therefore$ Perimeter of shaded region

= $AB+PB+\widehat{PA}$

= $r[tan\theta +rsec\theta -r+\frac{\pi r\theta }{{180}^{\circ }}]$

= $r[tan\theta +sec\theta +\frac{\pi \theta }{180}-1]$

Hence Proved.